Rút gọn: a) B = $\sqrt{3 \sqrt{2} \sqrt{3} \sqrt{6}.\sqrt{2}}$b) C = $\left(3 \sqrt{3}\right)\sqrt{2}.\sqrt{6-3\sqrt{3}}$c) D = $\left(\sqrt{3}-3\right)\sqrt{2}.\sqrt{6 3\sqrt{3}}$d) C = \(\sqrt...

NX

Nguyễn Xuân Nhi

17 tháng 7 2020 - olm

Rút gọn: a) B = $\sqrt{3+\sqrt{2}+\sqrt{3}+\sqrt{6}.\sqrt{2}}$b) C = $\left(3+\sqrt{3}\right)\sqrt{2}.\sqrt{6-3\sqrt{3}}$c) D = $\left(\sqrt{3}-3\right)\sqrt{2}.\sqrt{6+3\sqrt{3}}$d) C = $\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}$e) D...

Đọc tiếp

#Toán lớp 9

0

LL

Ly Ly

4 tháng 7 2021

* Rút gọn biểu...

Đọc tiếp

#Toán lớp 9

1

AT

An Thy

4 tháng 7 2021

a) $\dfrac{2\sqrt{125}-3\sqrt{5}-\sqrt{180}}{-\sqrt{5}}+\sqrt{8}=\dfrac{2\sqrt{25.5}-3\sqrt{5}-\sqrt{36.5}}{-\sqrt{5}}+\sqrt{8}$

$=\dfrac{10\sqrt{5}-3\sqrt{5}-6\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=\dfrac{\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=2\sqrt{2}-1$

b) $\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}$

$=\sqrt{3}-\sqrt{2}+3\sqrt{2}=2\sqrt{2}+\sqrt{3}$

c) $\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}=\sqrt{16.3}-2\sqrt{9.\dfrac{1}{3}}+\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}$

$=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}=1+\sqrt{3}$

d) $\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)$

$=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3$

Đúng(0)

LL

Ly Ly

7 tháng 7 2021

* Rút gọn các biểu thức

a. $\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{2\left(-5\right)^2}$

b. $\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-4}.\sqrt[3]{2}$

c. $6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-3\sqrt{8}$

d. $\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}-\dfrac{2}{\sqrt{3}-1}$

#Toán lớp 9

1

MY

missing you =

7 tháng 7 2021

a, $=>3-\sqrt{2}+\sqrt{50}=3-\sqrt{2}+5\sqrt{2}=3+4\sqrt{2}$

b, $=>\dfrac{\sqrt[3]{125.5}}{\sqrt[3]{5}}-\sqrt[3]{\left(-4\right).2}=\sqrt[3]{125}-\sqrt[3]{\left(-2\right)^3}$

$=5-\left(-2\right)=7$

c, $=>\sqrt{6}.\sqrt{\dfrac{6}{2}}-\sqrt{2}-3\sqrt{4.2}=\sqrt{6}.\sqrt{3}-\sqrt{2}-6\sqrt{2}$

$=\sqrt{18}-7\sqrt{2}=3\sqrt{2}-7\sqrt{2}=-4\sqrt{2}$

d, $=>\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{2}{\sqrt{3}-1}=\sqrt{3}-\dfrac{2}{\sqrt{3}-1}$

$=\dfrac{3-\sqrt{3}-2}{\sqrt{3}-1}=\dfrac{1-\sqrt{3}}{\sqrt{3}-1}=-1$

Đúng(1)

ND

nguyễn đăng khôi

30 tháng 5 2023

rút gọn A)$\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5-3}\right)^2}}$
B) $\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3+1}\right)^2}}}$
C) $\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}$

#Toán lớp 9

1

Y

YangSu

30 tháng 5 2023

$c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}$

$=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}$

$=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}$

$=\sqrt{29}$

Đúng(2)

NQ

Nhi Quỳnh

2 tháng 11 2023

Rút gọn các biểu thức saua) 2$\sqrt{32}$ + 3$\sqrt{72}-7\sqrt{50}+\sqrt{2}$ b)$\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}$ c) $\sqrt{11+6\sqrt{2}}-3+\sqrt{2}$d) $x-4+\sqrt{16-8x+x^2}\left(x4\right)$ e) \(\dfrac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}vớia...

Đọc tiếp

#Toán lớp 9

1

H9

HT.Phong (9A5)

CTVHS

2 tháng 11 2023

a) $2\sqrt{32}+3\sqrt{72}-7\sqrt{50}+\sqrt{2}$

$=2\cdot4\sqrt{2}+3\cdot6\sqrt{2}-7\cdot5\sqrt{2}+\sqrt{2}$

$=8\sqrt{2}+18\sqrt{2}-35\sqrt{2}+\sqrt{2}$

$=-8\sqrt{2}$

b) $\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}$

$=\left|3-\sqrt{3}\right|+\left|2-\sqrt{3}\right|$

$=3-\sqrt{3}+\sqrt{3}-2$

$=1$

c) $\sqrt{11+6\sqrt{2}}-3+\sqrt{2}$

$=\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-3+\sqrt{2}$

$=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}$

$=3+\sqrt{2}-3+\sqrt{2}$

$=2\sqrt{2}$

d) $x-4+\sqrt{16-8x+x^2}\left(x>4\right)$

$=x-4+\sqrt{x^2-8x+16}$

$=x-4+\sqrt{\left(x-4\right)^2}$

$=x-4+\left|x-4\right|$

$=x-4+x-4$

$=2x-8$

e) $\dfrac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\left(a< b\right)$

$=\dfrac{1}{a-b}\sqrt{\left[a^2\left(a-b\right)\right]^2}$

$=\dfrac{1}{a-b}\left|a^2\left(a-b\right)\right|$

$=\dfrac{-a^2\left(a-b\right)}{a-b}$

$=-a^2$

Đúng(0)

3T

30.Nɠυұễɳ Tɦàɲɦ Pɦúƈ ヅ

26 tháng 10 2023

Rút...

Đọc tiếp

#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

26 tháng 10 2023

a: $\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}$

$=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}$

$=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}$

$=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}$

b: $\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}$

$=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}$

$=\left|3+\sqrt{5}\right|+\left|3-\sqrt{5}\right|$

$=3+\sqrt{5}+3-\sqrt{5}=6$

c: $\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}$

$=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}$

$=2\sqrt{3}-3+2\sqrt{3}+3=4\sqrt{3}$

d: $\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}$

$=\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{\left(4-\sqrt{3}\right)^2}+3}$

$=\sqrt{\left(4+\sqrt{3}\right)\cdot\left(4-\sqrt{3}\right)+3}$

$=\sqrt{16-3+3}=\sqrt{16}=4$

e: $\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}$

$=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{3}$

$=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}$

Đúng(3)

B

bbiooo

10 tháng 1 2021

rút gọn các biểu thức sau:

a,$\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}$

b,$\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}$

c,$\sqrt{2+\sqrt{5-\sqrt{13-\sqrt{48}}}}$

d,$\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}$

#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

10 tháng 1 2021

a) Ta có: $\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}$

$=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}$

$=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}$

$=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}$

$=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}$

$=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}$

$=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}$

$=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}$

$=\sqrt{\sqrt{5}-\sqrt{5}+1}$

$=\sqrt{1}=1$

b) Ta có: $\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}$

$=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}$

$=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}$

$=\sqrt{6+2\sqrt{5}-\left(2\sqrt{5}-3\right)}$

$=\sqrt{6+3}=3$

c) Sửa đề: $\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}$

Ta có: $\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}$

$=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}$

$=\sqrt{2+\sqrt{5+\sqrt{\left(2\sqrt{3}-1\right)^2}}}$

$=\sqrt{2+\sqrt{5+2\sqrt{3}-1}}$

$=\sqrt{2+\sqrt{3+2\sqrt{3}\cdot1+1}}$

$=\sqrt{2+\sqrt{\left(\sqrt{3}+1\right)^2}}$

$=\sqrt{3+\sqrt{3}}$

d) Ta có: $\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}$

$=\dfrac{\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{2\sqrt{2}}$

$=\dfrac{\left(\sqrt{5}-1\right)^2\cdot\left(\sqrt{5}+1\right)+\left(\sqrt{5}+1\right)^2\cdot\left(\sqrt{5}-1\right)}{2\sqrt{2}}$

$=\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(\sqrt{5}-1+\sqrt{5}+1\right)}{2\sqrt{2}}$

$=\dfrac{4\cdot2\sqrt{5}}{2\sqrt{2}}$

$=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}$

Đúng(3)

NT

Nguyễn Thị Yến

5 tháng 10 2021

rút gọn

a) $\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$

b) $\left(\sqrt{7-3\sqrt{5}}\right)\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)$

c) $\left(\sqrt{14}-\sqrt{10}\right)\left(6-\sqrt{35}\right)\left(\sqrt{6+\sqrt{35}}\right)$

#Toán lớp 9

2

NL

Nguyễn Lê Phước Thịnh

5 tháng 10 2021

b: Ta có: $\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)$

$=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)$

$=4\left(7+3\sqrt{5}\right)$

$=28+12\sqrt{5}$

Đúng(2)

AH

Akai Haruma

Giáo viên

5 tháng 10 2021

Lời giải:

a.

$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$

$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$

$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$

$=2\sqrt{5}-5\sqrt{10}$

$\Rightarrow A=\sqrt{10}-5\sqrt{5}$

b.

$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$

$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$

$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$

$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$

$\Rightarrow B=28+12\sqrt{5}$

c.

$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$

$=(7-5)(6-\sqrt{35})$

$=2(6-\sqrt{35})=12-2\sqrt{35}$

Đúng(1)

QT

Quyết Tâm Chiến Thắng

17 tháng 9 2019 - olm

Rút gọn...

Đọc tiếp

#Toán lớp 9

4

AN

alibaba nguyễn

18 tháng 9 2019

d/ $x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}$

$\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}$

$\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}$

$\Leftrightarrow x^3=6-5x$

$\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0$

$\Leftrightarrow x=1$

Đúng(0)

AN

alibaba nguyễn

19 tháng 9 2019

c/

$\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}$

$=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}$

$=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}$

$=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}$

$=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}$

$=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}$

$=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}$

$=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}$

$=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}$

$=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)$

$=3-1=2$

Đúng(0)

QE

Quynh Existn't

26 tháng 6 2021

...

Đọc tiếp

#Toán lớp 9

2

NN

Nguyễn Ngọc Lộc

26 tháng 6 2021

$A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|$

$=54+8-32=30$

$B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}$

$=5-2\sqrt{2}$

$C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}$

$=2-2\sqrt{3}$

$D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}$

$=2\sqrt{6}$

$E=\sqrt{15^2}-\sqrt{5^2}=15-5=10$

Đúng(2)

Y

Yeutoanhoc

26 tháng 6 2021

`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`

Đúng(1)

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP